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Linear Algebra: Coordinate system corresponding to the basis

  • Thread starter RyanV
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  • #1
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Homework Statement


In the xy-plane, sketch the coordinate system [ a; b] corresponding to the basis
{ (1, 1 ) , (1, -1) } by drawing the lines a = 0, [tex]\pm[/tex]1 and b = 0, [tex]\pm[/tex]1. What point in the xy-plan corresponds to a = 1, b = 2?


Homework Equations


Not sure of any in this case


The Attempt at a Solution


I just stared at this Q in total blank.
After couple minutes, I tried to place the a and b coordinates like they were x and y coordinates respectively, and then line them up according to which one was a = 0, a = 1. Once again, had no clue what I was doing.

Had a look at the answer provided, and it looked like the attached provided. Tried to make heads and tails of it but doesn't really make much sense to me..

Any assistance would be greatly appreciated! =D
 

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Answers and Replies

  • #2
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A basis a linear combination of vectors which are linearly independent (meaning if you have n of them, you cannot add/subtract n-1 of them to get the last one) and which span (can be used to define) a given space. In this case we have R(2) (or E(2), depending on whatever notation you use).

Normally in R(2) we use the simple I (1,0) and J (0,1) vectors, which are linearly independent, but now we want to use a different system, (a,b) which also "spans" R(2) (dot (1, 1 ) , (1, -1) and it will equal zero). So we can relate these two systems because they span the same space.

x(1,0)+y(0,1)=a(1,1)+b(1,-1)

a=1, b=2
1(1,1)+2(1,-1) = (1+2, 1-2)= (3,-1) = x(1,0)+ y(0,1)

we can easily see x will be the first part of the vector (3,-1) and y will be the second part, so the (x,y) = (3,-1)

hope that helped!
 
  • #3
12
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ahh, I see. thanks...that part of the question does make sense now.

But i'm still unsure of the drawing of the lines...If you have a look at the lines that are drawn in the answer, how did they come to that? I'm thinking of the basis that was given, but from where I stand, it only explains the lines that kinda go y = [tex]\pm[/tex] x since B = { (1,1), (-1,1) }.
Unless of course, my thinking is not correct.

Thanks again =)
 

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